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Consider two solid spheres $\mathrm{P}$ and $\mathrm{Q}$ each of density $8 \mathrm{gm} \mathrm{cm}^{-3}$ and diameters $1 \mathrm{~cm}$ and $0.5 \mathrm{~cm}$, respectively. Sphere $\mathrm{P}$ is dropped into a liquid of density $0.8 \mathrm{gm} \mathrm{cm}^{-3}$ and viscosity $\eta=3$ poiseulles. Sphere $Q$ is dropped into a liquid of density $1.6 \mathrm{gm} \mathrm{cm}^{-3}$ and viscosity $\eta=2$ poiseulles. The ratio of the terminal velocities of $\mathrm{P}$ and $\mathrm{Q}$ is
$4$
$2$
$1$
$3$
Solution
$\text { Terminal velocity } v =\frac{2 gr ^2(\rho-\sigma)}{9 \eta}$
$\Rightarrow \frac{v_p}{v_Q}=\frac{r_p^2\left(\rho_p-\sigma_1\right) \eta_2}{r_Q^2\left(\rho_Q-\sigma_2\right) \eta_1}$
Given: $\eta_1=3 PI =30 P \eta_2=2 PI =20 P \rho_{ p }=\rho_{ Q }=8 gcm ^{-3} \sigma_1=0.8$ $gcm ^{-3} \sigma_2=1.6 \ gcm ^{-3}$
Radius of sphere $\operatorname{Pr}_{ p }=\frac{1}{2} cm=0.005 \ m$
Radius of sphere $Q r _{ Q }=\frac{0.5}{2} cm=0.0025 \ m$
$\therefore \frac{v_p}{v_Q}=\frac{(0.005)^2(8-0.8)(20)}{(0.0025)^2(8-1.6)(30)}=3$
